Problem: $ F = \left[\begin{array}{rr}-2 & 0 \\ 1 & 1\end{array}\right]$ $ E = \left[\begin{array}{rr}4 & -2 \\ 3 & 5\end{array}\right]$ What is $ F E$ ?
Solution: Because $ F$ has dimensions $(2\times2)$ and $ E$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ F E = \left[\begin{array}{rr}{-2} & {0} \\ {1} & {1}\end{array}\right] \left[\begin{array}{rr}{4} & \color{#DF0030}{-2} \\ {3} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{4}+{0}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{4}+{0}\cdot{3} & ? \\ {1}\cdot{4}+{1}\cdot{3} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{4}+{0}\cdot{3} & {-2}\cdot\color{#DF0030}{-2}+{0}\cdot\color{#DF0030}{5} \\ {1}\cdot{4}+{1}\cdot{3} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{4}+{0}\cdot{3} & {-2}\cdot\color{#DF0030}{-2}+{0}\cdot\color{#DF0030}{5} \\ {1}\cdot{4}+{1}\cdot{3} & {1}\cdot\color{#DF0030}{-2}+{1}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-8 & 4 \\ 7 & 3\end{array}\right] $